Question

An open rectangular tank 1.5 m wide 2m deep and 2m long is half filled with water. It is accelerated horizontally at 3.27 m/sec2 in the direction of its length. Determine the depth of water at each end of tank. [g = 9.81 m/sec2 ]

Answer 1

The depths of water at the two ends of the tank are $\frac{2}{3}$ m and $\frac{4}{3}$ m.

Answer 2

In an accelerating frame of reference, the free surface of a liquid remains perpendicular to the effective gravitational force. The effective gravitational acceleration is $\vec{g}_{eff} = \vec{g} - \vec{a}$. The slope of the free surface in the direction of acceleration ($x$) is given by $\frac{dz}{dx} = \frac{a_x}{g}$.

Given $a_x = 3.27$ m/s² and $g = 9.81$ m/s², the slope is $m = \frac{3.27}{9.81} = \frac{1}{3}$.

The free surface can be described by the equation $z(x) = z_1 + mx$, where $z_1$ is the depth at one end of the tank (say, $x=0$), and $x$ is the distance along the length of the tank. So, $z(x) = z_1 + \frac{1}{3}x$.

The volume of water in the tank remains constant. Initial volume $V = \text{Length} \times \text{Width} \times \text{Initial Depth} = 2 \text{ m} \times 1.5 \text{ m} \times 1 \text{ m} = 3 \text{ m}^3$.

The volume of water in the tilted tank is given by the integral of the depth over the length: $V = \int_0^L W z(x) \, dx = W \int_0^L \left(z_1 + \frac{1}{3}x\right) \, dx$. $V = W \left[z_1 x + \frac{1}{3} \frac{x^2}{2}\right]_0^L = W \left(z_1 L + \frac{L^2}{6}\right)$.

Substitute the known values: $V=3$ m³, $W=1.5$ m, $L=2$ m. $3 = 1.5 \left(z_1 \times 2 + \frac{2^2}{6}\right)$ $3 = 1.5 \left(2z_1 + \frac{4}{6}\right)$ $3 = 1.5 \left(2z_1 + \frac{2}{3}\right)$ Divide by 1.5: $2 = 2z_1 + \frac{2}{3}$ $2z_1 = 2 - \frac{2}{3} = \frac{6-2}{3} = \frac{4}{3}$ $z_1 = \frac{2}{3}$ m.

This is the depth at one end ($x=0$). The depth at the other end ($x=L=2$) is: $z_2 = z_1 + \frac{1}{3}L = \frac{2}{3} + \frac{1}{3}(2) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$ m.

The depths at the two ends of the tank are $\frac{2}{3}$ m and $\frac{4}{3}$ m.